Question

In the following exercises, solve by using the Quadratic Formula:$t^2+13t=-40$

Short answer

The solution of the quadratic equation is$t=-5;t=-8$

Step-by-step solution

Given information

We are given a quadratic equation$t^2+13t=-40$

We add 40 on both sides and the identify the values of a,b,c

On adding $40$on both sides we get

$t^2+13t+40=0$

On comparing with standard equation we get,$a=1,b=13,c=40$

Write the quadratic formula and the substitute the values in it

The quadratic formula can be given as

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

On substituting the values we get

$t=\frac{-13\pm \sqrt{169-4\left(1\right)\left(40\right)}}{2\left(1\right)}$

Simplify

On simplifying we get

$$\begin{gathered} t=\frac{-13\pm \sqrt{169-160}}{2} \\ t=\frac{-13\pm \sqrt{9}}{2} \\ t=\frac{-13\pm 3}{2} \end{gathered}$$

On further simplifying we get

$$\begin{gathered} t=\frac{-13+3}{2} \\ t=\frac{-10}{2} \\ t=-5 \end{gathered}$$and$$\begin{gathered} t=\frac{-13-3}{2} \\ t=\frac{-16}{2} \\ t=-8 \end{gathered}$$

Conclusion

The solution of the above equation is$q=-5,q=-8$